Check if 2 * K + 1 non-empty strings exists whose concatenation forms the given string

Given a string S consisting of N characters and positive integer K, the task is to check if there exist any (K + 1) strings i.e., A₁, A₂, A₃, …, A_K, A_{(K + 1)} such that the concatenation of strings  A₁, A₂, A₃, …, A_K, and A_{(K + 1)} and the concatenation of the reverse of each strings A_K, A_{(K – 1)}, A_{(K – 2)}, …, A₁, and A₀ is the string S. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: S = “qwqwq”, K = 1
Output: Yes
Explanation:
Consider the string A₁ as “qw”, and A₂ as “q”. Now the concatenation of  A₁, A₂, reverse of  A₁ is “qwqwq”, which is the same as the given string S.

Input: S = “qwqwa”, K = 2
Output: No

Approach: The given problem can be solved based on the observation that for a string S to satisfy the given condition, the first K characters must be equal to the last K characters of the given string. Follow the steps below to solve the problem:

• If the value of (2*K + 1) is greater than N, then print “No” and return from the function.
• Otherwise, store the prefix of size K i.e., S[0, …, K] in a string A, and the suffix of size K i.e., S[N – K, …, N – 1] in a string B.
• Reverse the string B and check if A is equal to B or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if the string S
// can be obtained by (K + 1) non-empty
// substrings whose concatenation and
// concatenation of the reverse
// of these K strings
void checkString(string s, int k)
{
    // Stores the size of the string
    int n = s.size();

    // If n is less than 2*k+1
    if (2 * k + 1 > n) {
        cout << "No";
        return;
    }

    // Stores the first K characters
    string a = s.substr(0, k);

    // Stores the last K characters
    string b = s.substr(n - k, k);

    // Reverse the string
    reverse(b.begin(), b.end());

    // If both the strings are equal
    if (a == b)
        cout << "Yes";
    else
        cout << "No";
}

// Driver Code
int main()
{
    string S = "qwqwq";
    int K = 1;
    checkString(S, K);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{

      // Function to check if the string S
    // can be obtained by (K + 1) non-empty
    // substrings whose concatenation and
    // concatenation of the reverse
    // of these K strings
    static void checkString(String s, int k)
    {

        // Stores the size of the string
        int n = s.length();

        // If n is less than 2*k+1
        if (2 * k + 1 > n) {
            System.out.println("No");
            return;
        }

        // Stores the first K characters
        String a = s.substring(0, k);

        // Stores the last K characters
        String b = s.substring(n - k, n);

        // Reverse the string
        StringBuffer str = new StringBuffer(b);
        // To reverse the string
        str.reverse();
        b = str.toString();

        // If both the strings are equal
        if (a.equals(b))
            System.out.println("Yes");
        else
            System.out.println("No");
    }

    // Driver Code
    public static void main (String[] args)
    {
        String S = "qwqwq";
        int K = 1;
        checkString(S, K);
    }
}

// This code is contributed by Dharanendra L V.

Python3

# Python3 program for the above approach

# Function to check if the S
# can be obtained by (K + 1) non-empty
# substrings whose concatenation and
# concatenation of the reverse
# of these K strings
def checkString(s, k):

    # Stores the size of the string
    n = len(s)

    # If n is less than 2*k+1
    if (2 * k + 1 > n):
        print("No")
        return

    # Stores the first K characters
    a = s[0:k]

    # Stores the last K characters
    b = s[n - k:n]

    # Reverse the string
    b = b[::-1]

    # If both the strings are equal
    if (a == b):
        print("Yes")
    else:
        print("No")

# Driver Code
if __name__ == '__main__':
    S = "qwqwq"
    K = 1
    checkString(S, K)

# This code is contributed by mohit kumar 29.

C#

// C# program for the above approach
using System;
class GFG {

    // Function to check if the string S
    // can be obtained by (K + 1) non-empty
    // substrings whose concatenation and
    // concatenation of the reverse
    // of these K strings
    static void checkString(string s, int k)
    {

        // Stores the size of the string
        int n = s.Length;

        // If n is less than 2*k+1
        if (2 * k + 1 > n) {
            Console.Write("No");
            return;
        }

        // Stores the first K characters
        string a = s.Substring(0, k);

        // Stores the last K characters
        string b = s.Substring(n - k, k);

        // Reverse the string
        char[] arr = b.ToCharArray();
        Array.Reverse(arr);
        b = new String(arr);

        // If both the strings are equal
        if (a == b)
            Console.Write("Yes");
        else
            Console.Write("No");
    }

    // Driver Code
    public static void Main()
    {
        string S = "qwqwq";
        int K = 1;
        checkString(S, K);
    }
}

// This code is contributed by ukasp.

Javascript

<script>
// Javascript program for the above approach


// Function to check if the string S
// can be obtained by (K + 1) non-empty
// substrings whose concatenation and
// concatenation of the reverse
// of these K strings
function checkString(s, k)
{
    // Stores the size of the string
    let n = s.length;

    // If n is less than 2*k+1
    if (2 * k + 1 > n) {
        document.write("No");
        return;
    }

    // Stores the first K characters
    let a = s.substr(0, k);

    // Stores the last K characters
    let b = s.substr(n - k, k);

    // Reverse the string
    b.split("").reverse().join("")

    // If both the strings are equal
    if (a == b)
        document.write("Yes");
    else
        document.write("No");
}

// Driver Code
    let S = "qwqwq";
    let K = 1;
    checkString(S, K);

// This code is contributed by gfgking.
</script>

Output:

Yes

Time complexity: O(N) 
Auxiliary Space: O(N)